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Disprove that if $n$ is an odd positive integer greater than $1$ then $2^n-1$ is prime.

So my approach to this is as follows but I have no idea where to go from here or whether this is the right approach.

Let $n=2x+1$ where $x\in\Bbb Z^+$

Then $2^n+1=2^{2x+1}-1$

$=6(2^n)-1$

H.Linkhorn
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  • All you need is a single counterexample, and there is a fairly small one. – lulu Dec 16 '18 at 19:32
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    is there not a way this can be proved by deduction then? – H.Linkhorn Dec 16 '18 at 19:32
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    You are trying to disprove something. In truth, there are primes of this form and there are composites of this form. To disprove this claim you just need to find a single composite example. – lulu Dec 16 '18 at 19:34
  • If you want a harder problem, you can try to prove that there are infinitely many composites of this form. – lulu Dec 16 '18 at 19:35
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    Oh okay, the question is worth 4 marks and i thought a simple counter example way to simple – H.Linkhorn Dec 16 '18 at 19:35
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    Nope. All it takes is one counterexample to disprove a theorem. – lulu Dec 16 '18 at 19:35

2 Answers2

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In general:

  • If $n$ is composite, then $2^n-1$ is composite. The prove is not hard to follow up. since $n$ is composite, $n = km$. This implies that $2^{km}-1 = (2^k-1)(1+2^k+2^{2k}+2^{3k}...+2^{km-1})$.

  • If $2^n - 1$ is prime, then $n$ is prime. The prove can be deduced from the previous part by "counter-positiving" this statement. "NOTE THAT THE CONVERSE IS NOT TRUE".

A quick counter example to disprove, take $n$ to be $9$, then $ 2^9-1 = 511 = 7*73$ which is divisible by $2^3 - 1 = 7$ and $1+2^3+2^6 = 73$.

Maged Saeed
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Take $$2^{11}-1=23\cdot 89$$ and this number is not prime

Dr. Sonnhard Graubner
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